Problem: Let $f(x)=\dfrac{\cos(x)}{x}$. $f'(x)=$
Answer: $f(x)$ is the quotient of two, more basic, expressions: $\cos(x)$ and $x$. Therefore, the derivative of $f$ can be found using the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! $\begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}\left[\dfrac{\cos(x)}{x}\right] \\\\ &=\dfrac{\dfrac{d}{dx}[\cos(x)]x-\cos(x)\dfrac{d}{dx}[x]}{x^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{-\sin(x)\cdot x-\cos(x)\cdot1}{x^2}&&\gray{\text{Differentiate }\cos(x)\text{ and }x} \\\\ &=-\dfrac{\cos(x)+x\sin(x)}{x^2}&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $f'(x)=-\dfrac{\cos(x)+x\sin(x)}{x^2}$ or any other equivalent form.